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Question Number 154651 by mathdanisur last updated on 20/Sep/21

let be A = ((1,1),(0,1) ) ; B = ((1,0),(1,1) ) find š›€ = e^A āˆ™ (e^B )^(āˆ’1) (e^A - exponential matrix)

$$\mathrm{let}\:\mathrm{be}\:\:\boldsymbol{\mathrm{A}}\:=\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\:\:;\:\:\boldsymbol{\mathrm{B}}\:=\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{find}\:\:\boldsymbol{\Omega}\:=\:\mathrm{e}^{\boldsymbol{\mathrm{A}}} \:\centerdot\:\left(\mathrm{e}^{\boldsymbol{\mathrm{B}}} \right)^{āˆ’\mathrm{1}} \\ $$$$\left(\mathrm{e}^{\boldsymbol{\mathrm{A}}} \:-\:\mathrm{exponential}\:\mathrm{matrix}\right) \\ $$

Answered by TheHoneyCat last updated on 20/Sep/21

let nāˆˆN ((1,n),(0,1) )Ɨ ((1,1),(0,1) )= ((1,(1+n)),(0,1) ) so by a trivual recurence āˆ€nāˆˆN A^n = ((1,n),(0,1) ) by definition e^A =Ī£_(nāˆˆN) (1/(n!))A^n so e^A = ((e,e),(0,e) )=e.A and since B=^t A, e^B =e.^t A hence š›€=(e/e)A(^t A)^(āˆ’1) = ((1,1),(0,1) )Ɨ ((1,0),((āˆ’1),1) ) = ((0,1),((āˆ’1),1) )_ā– 

$$\mathrm{let}\:{n}\in\mathbb{N} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}Ɨ\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}+{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{so}\:\mathrm{by}\:\mathrm{a}\:\mathrm{trivual}\:\mathrm{recurence} \\ $$$$\forall{n}\in\mathbb{N}\:\boldsymbol{\mathrm{A}}^{{n}} =\begin{pmatrix}{\mathrm{1}}&{{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{by}\:\mathrm{definition}\:{e}^{\boldsymbol{\mathrm{A}}} =\underset{{n}\in\mathbb{N}} {\sum}\frac{\mathrm{1}}{{n}!}\boldsymbol{\mathrm{A}}^{{n}} \\ $$$$\mathrm{so}\:{e}^{\boldsymbol{\mathrm{A}}} =\begin{pmatrix}{{e}}&{{e}}\\{\mathrm{0}}&{{e}}\end{pmatrix}={e}.\boldsymbol{\mathrm{A}} \\ $$$$\mathrm{and}\:\mathrm{since}\:\boldsymbol{\mathrm{B}}=^{{t}} \boldsymbol{\mathrm{A}},\:{e}^{\boldsymbol{\mathrm{B}}} ={e}.^{{t}} \boldsymbol{\mathrm{A}} \\ $$$$\mathrm{hence}\:\boldsymbol{\Omega}=\frac{{e}}{{e}}\boldsymbol{\mathrm{A}}\left(^{{t}} \boldsymbol{\mathrm{A}}\right)^{āˆ’\mathrm{1}} \\ $$$$=\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}Ɨ\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{āˆ’\mathrm{1}}&{\mathrm{1}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\mathrm{0}}&{\mathrm{1}}\\{āˆ’\mathrm{1}}&{\mathrm{1}}\end{pmatrix}_{\blacksquare} \\ $$

Commented by mathdanisur last updated on 20/Sep/21

Very nice Ser, thankyou

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{Ser},\:\mathrm{thankyou} \\ $$

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